Sample Problem 4.5 A 400-lb weight is attached at A to the lever shown. The constant of the spring BC is k = 250 lb/in., and the spring is unstretched when e = 0. Determine the position of equilibrium. 1=8 in. k= 250 lb/in. r= 3 in. W = 400 Ib STRATEGY: Draw a free-body diagram of the lever and cylinder to show all forces acting on the body (Fig. 1), then sum moments about O. Your final answer should be the angle 0. MODELING: Free-Body Diagram. Denote by s the deflection of the spring from its unstretched position and note that s = re. Then F = ks = kre. ANALYSIS: Equilibrium Equation. Sum the moments of W and F about O to eliminate the reactions supporting the cylinder. The result is kr2 + 1ΣΜο-0: WI sin 0 – r(kre) = 0 sin 0 WI Substituting the given data yields (250 lb/in.)(3 in.)², sin 0 -0 sin e = 0.703 0 (400 Ib)(8 in.) Solving by trial and error, the angle is e = 0 e = 80.3° REFLECT and THINK: The weight could represent any vertical force acting on the lever. The key to the problem is to express the spring force as a function of the angle 0. Unstretched position -I sin 0 F=ks R, Fig. 1 Free-body diagram of the lever and cylinder.

Sample Problem 4.5 A 400-lb weight is attached at A to the lever shown. The constant of the spring BC is k = 250 lb/in., and the spring is unstretched when e = 0. Determine the position of equilibrium. 1=8 in. k= 250 lb/in. r= 3 in. W = 400 Ib STRATEGY: Draw a free-body diagram of the lever and cylinder to show all forces acting on the body (Fig. 1), then sum moments about O. Your final answer should be the angle 0. MODELING: Free-Body Diagram. Denote by s the deflection of the spring from its unstretched position and note that s = re. Then F = ks = kre. ANALYSIS: Equilibrium Equation. Sum the moments of W and F about O to eliminate the reactions supporting the cylinder. The result is kr2 + 1ΣΜο-0: WI sin 0 – r(kre) = 0 sin 0 WI Substituting the given data yields (250 lb/in.)(3 in.)², sin 0 -0 sin e = 0.703 0 (400 Ib)(8 in.) Solving by trial and error, the angle is e = 0 e = 80.3° REFLECT and THINK: The weight could represent any vertical force acting on the lever. The key to the problem is to express the spring force as a function of the angle 0. Unstretched position -I sin 0 F=ks R, Fig. 1 Free-body diagram of the lever and cylinder.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Solve Sample Prob. 4.5, assuming that the spring is unstretched when 0 = 90°.
(Reference to Sample Problem 4.5):

Sample Problem 4.5
A 400-lb weight is attached at A to the lever shown. The constant of the
spring BC is k = 250 lb/in., and the spring is unstretched when e = 0.
Determine the position of equilibrium.
1=8 in.
k= 250 lb/in.
r= 3 in.
W = 400 Ib
STRATEGY: Draw a free-body diagram of the lever and cylinder to
show all forces acting on the body (Fig. 1), then sum moments about O.
Your final answer should be the angle 0.
MODELING:
Free-Body Diagram. Denote by s the deflection of the spring from
its unstretched position and note that s = re. Then F = ks = kre.
ANALYSIS:
Equilibrium Equation. Sum the moments of W and F about O to
eliminate the reactions supporting the cylinder. The result is
kr2
+ 1ΣΜο-0:
WI sin 0 – r(kre) = 0
sin 0
WI
Substituting the given data yields
(250 lb/in.)(3 in.)²,
sin 0
-0 sin e = 0.703 0
(400 Ib)(8 in.)
Solving by trial and error, the angle is
e = 0
e = 80.3°
REFLECT and THINK: The weight could represent any vertical force
acting on the lever. The key to the problem is to express the spring force
as a function of the angle 0.
Unstretched
position
-I sin 0
F=ks
R,
Fig. 1 Free-body diagram of the lever
and cylinder.

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